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I'm trying to implement a high-side DC switching device (60 V, a couple of amps) using an N-channel MOSFET driven by a IRS20752L high-side gate driver.

The IRS20752L datasheet gives a typical application schematic (below) which I followed but there is no mention of any cap values.

As I understand it, the cap between Vb and Vs would be the bootstrap cap and the one between Vcc and COM would be a decoupling cap. Am I correct on that?

Now, I do not have any remote idea as to what values they should be.

How do I calculate the values for these capacitors?

enter image description here

Here is a schematic draft : enter image description here

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    \$\begingroup\$ What is the load? Note that this device does not include a charge pump, so it depends upon the load always going to ~0V on a regular basis. If you cannot ensure this, you must add a charge pump or isolated supply (to VB) externally, or choose a different type which does. (This is avoided in typical SMPS applications where a dual driver and half-bridge ensure the output pulls low between cycles.) \$\endgroup\$
    – Tim Williams
    Mar 5 at 2:23
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    \$\begingroup\$ @TimWilliams Thanks for the answer. The driver/mosfet is meant to switch a DC load in a "SSPC" / relay kind of way so not going to 0V on a regular basis... so that might be a problem... I've added a schematic draft hoping it might clear things up. But how does such a gate driver work without a charge pump? \$\endgroup\$
    – VoltsAndNuts
    Mar 5 at 11:26
  • \$\begingroup\$ You may also want a flyback diode, if the load can be inductive. Such a gate driver, without regularly returning VS near GND, and without a charge pump, doesn't work. For example if set high (100% duty cycle) for some time, the bootstrap capacitor discharges until UVLO, then it stays latched off until the next CONTROL rising edge AND high-side voltage is adequate. Also note that the fuse protects wiring, not the electronics (it will not protect the MOSFET, in case that was your intention). \$\endgroup\$
    – Tim Williams
    Mar 5 at 16:16
  • \$\begingroup\$ @TimWilliams thank you for the clarification. As clear as it is unfortunate since the PCB is already underway. I'm a bit lost as to what type of gate-driver I need... What keywords / type should I look for? \$\endgroup\$
    – VoltsAndNuts
    Mar 5 at 19:42
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    \$\begingroup\$ Perhaps you should start a new question, starting with what type and range of expected loads you are driving, at what frequency/speed, power level, etc., and asking for suggestions from there. Whether that can be bodged onto your "underway" board is another matter. Again, if nothing else, an isolated supply can be wired to [VB, VS], which you can almost certainly find a way to bodge in. \$\endgroup\$
    – Tim Williams
    Mar 5 at 19:49

2 Answers 2

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The bootstrap cap must provide enough charge to turn on the MOSFET gate, without dropping its own voltage significantly. It also must be close to the IC because it forms a very high speed current loop. That usually means that at least 10x the input capacitance of the FET is required, in a small package. 100 nF is a value I see often. It can be backed up with a higher resistance bulk cap as suggested by @TonyM. This helps in situations where you want very long on-times. I think it is not good to use another low ESR (e.g. MLCC) cap for this bulk storage as that would resonate with the 100 nF cap closer to the IC.

The input cap is probably less critical in a pure high-side driver as it only backs up the input logic. So it makes sure the logic levels are stable and don't give rise to unintentional toggling. 100 nF is also a sensible value here, but 10 nF would be sufficient I guess.

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This Infineon IRS20752L application note, DT04-04 'Using Monolithic High Voltage Gate Drivers', gives in detail the procedure and equations for calculating the bootstrap capacitor value (introductory sample show below).

It refers to the supply decoupling capacitance but does not specify how to calculate it. You could use very a rough estimate value for the decoupling capacitor, such as 100 nF in parallel with 10 uF, located close to the component pins with the 100 nF closest. A calculation is better but these values are likely to be plenty. They're much higher than the Ciss (gate-source capacitance) of even a large MOSFET.

(START OF APP NOTE, READ IT FURTHER FOR DETAILED SOLUTION)

enter image description here

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  • \$\begingroup\$ am I correct in assuming I can use 20V (Vcc to Com being 12V) caps? 47µF / 20V will already be quite big so if I start looking for 47µF / 60V, the package start being huge... \$\endgroup\$
    – VoltsAndNuts
    Mar 5 at 11:29
  • \$\begingroup\$ @VoltsAndNuts, have revised this answer. \$\endgroup\$
    – TonyM
    Mar 5 at 15:36

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